Extracting the Root Element from the XML File

Shell Programming and Scripting
1

Any help to extract the root element from an XML file will be appreciated.

Example: test.xml

<?xml version="1.0" encoding="utf-8" ?> 
<TestXMLMessage>
<TestRec>
<ID>1000</ID> 
</TestRec>
</TestXMLMessage>

Wanted to extract the TestXMLMessage.

Regards,

Chari

2

Something like this?

awk -F"[<>]" 'NR==2{print $2}' file

or maybe this what you mean:

awk '/<TestXMLMessage>/{f=1};f;/<\/TestXMLMessage>/{f=0}' file
1 Like
3

Could this help you?

awk '{if(/^<TestXMLMessage>/){print;flg=1;next} if(flg==1) {print;next} if (/^<\/TestXMLMessage>/){print;flg=0}}' test.xml
4

This can be done a bit shorter:

awk '/<TestXMLMessage>/,/<\/TestXMLMessage>/' file
1 Like
5

Here is an XSL stylesheet which prints the root element of the document

<xsl:stylesheet version="1.0"
   xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

   <xsl:output method="text"/>

   <xsl:template match="/">
      <xsl:value-of select="name(/*)"/>
   </xsl:template>

</xsl:stylesheet>

Using xsltproc to transform the input docuemnt:

$ xsltproc example.xsl example.xml
TestXMLMessage
$
6

Hi All,

Thanks for your quick response.

This helped what I wanted to achieve.

awk -F"[<>]" 'NR==2{print $2}' file

Best Regards,

Chari

7

Hi All,

I am trying to change the XML filename in a folder to the rootelement in the XML file.

Now I have something like this.

for i in *.xml; do echo $i;awk -F"[<>]" 'NR==2{print $2}' $i; done

Output of above command:

TestMsg2010-10-19_20_20_54.xml
R1Msg TimeStamp="2010-10-19T08:49:08.000000Z"
TestMsg2010-10-19_20_20_58.xml
R2Msg TimeStamp="2010-10-19T08:49:08.000000Z"

Renamed File Output:(This is what I want to achieve)

R1Msg2010-10-19_20_20_54.xml
R2Msg2010-10-19_20_20_58.xml
R3Msg2010-10-19_20_21_00.xml

Is there a way that I can tweak the above command to achieve the same? Please help me as I am new to this side of programming.

Thanks in advance.